Moment of Inertia

As stated in the article quoted by Sparkstation, Moment of Inertia of any section is equal to the integral of each element of area times the square of the distance to the center of gravity.

For the square tube, neglecting the rounded corners, you could consider the area as the sum of two squares, the smaller one a negative area, representing the square hole. For the outer square, Io = d⁴/12 where d is the outer dimension of the tube. For the inner square, Ii = (d-2t)⁴/12 where t is the thickness of the tube. Then, for the square tube, I = Io - Ii.

For the angle, you must determine the location of the centroid of the section, then calculate the moment of inertia of each rectangle about its own center of gravity and add a correction of A*y² where y is the distance from the centroid of the rectangle to the centroid of the section. So I = ∑ I(self) + ∑ A*y².

If you wish to consider the rounded corners of the square tube and the fillet of the angle, you will have to do a little more work.